Ta có:
\(I=\int \frac{{\rm d}x}{\sin x.\sin \left(x+\frac{\pi }{6} \right)} =\int \frac{{\rm d}x}{\sin x.\left(\frac{\sqrt{3} }{2} \sin x+\frac{1}{2} \cos x\right)}\)
\(=\int \frac{{\rm d}x}{\sin ^{2} x.\left(\frac{\sqrt{3} }{2} +\frac{1}{2} \cot x\right)} .\)
Đặt \(t=\frac{\sqrt{3} }{2} +\frac{1}{2} \cot x\Rightarrow {\rm d}t=-\frac{1}{2} .\frac{1}{\sin ^{2} x} {\rm d}x\Rightarrow \frac{1}{\sin ^{2} x} {\rm d}x=-2{\rm d}t.\)
Suy ra: \(I=\int \frac{-2{\rm d}t}{t} =-2\ln \left|t\right|+C. Vậy I=-2\ln \left|\frac{\sqrt{3} }{2} +\frac{1}{2} \cot x\right|+C.\)